- #1

PingPong

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## Homework Statement

Prove that every compact set is bounded.

## Homework Equations

The usual compactness stuff - a compact set in a metric space X is one that, for every open cover, there is a finite subcover.

## The Attempt at a Solution

I'm really hesitant about this question because my professor kept repeating that there is much more to compactness in general metric spaces than there was in real analysis (where compact sets are closed and bounded). A proof that I've come up with is essentially the same proof used for the Heine-Borel theorem, and I don't think it works.

Suppose a set E is compact, and consider a neighborhood around a point p, [itex]I_n=N_n(p)[/itex]; of course, [itex]\{I_n\}[/itex] serves as a cover for E because there exists an n such that every q in X is also in [itex]N_n(p)[/itex]*. But E is compact, so there a finite subcover [itex]\{I_{n_k}\}[/itex], which implies that E is bounded.

My problem with that is with the part marked *. Isn't this already assuming that it's bounded? It works for real numbers because of the Archimedian property, but a general metric space doesn't have this property.

Of course, I could always try the opposite and show that if a set E is not bounded, then it is not compact, right? For example, if I used the metric [itex]d(x,y)=\infty[/itex] if [itex]x\ne y[/itex], then this metric on any set makes the set unbounded. Then it boils down to finding a cover that has no finite subcover...such as, perhaps, [itex]I_n=\{N_n(p):n\in N\}[/itex] for some [itex]p\in E[/itex].

Am I anywhere near close or am I just off my rocker?